1
$\begingroup$

Let $R$, $S$ be rings with identity. A map $f: R \times R \to S$ is said to be an a $R_S$-Pascal map if, for all $r_1, r_2 \in R$, the following relations are satisfied : $$\begin{align*} f(r_1-1_G, &r_2-1_R)+f(r_1-1_R,r_2)=f(r_1,r_2) \\ &f(r_1,r_2) = f(r_1,r_1-r_2). \end{align*}$$

Example of Pascal maps include the trivial $R_S$-Pascal map $f(r_1,r_2) = 0_R$ for all $r_1,r_2 \in R$, the $\mathbb{Z}_\mathbb{Z}$-Pascal map defined by $$f(n,k)= \binom{n}{k},$$ for $0 \leq k \leq n$ and $0$ otherwise, and the $\mathbb{Z}_\mathbb{Z}$-Pascal map defined by $$f(n,k)= \binom{n}{k}_S,$$ for $0 \leq k \leq n$ and $0$ otherwise, where $\displaystyle \binom{n}{k}_S = \frac{n!_S}{k!_S(n-k)!_S}$ is the Bhargava binomial coefficient over a set $S$. (Note that $1_R$ is not an $R_G$-Pascal map unless the ring $R$ has characteristic 1).

Let $\mathrm{Bin}_R(S)$ denote the collection of all $R_S$-Pascal maps; then $\mathrm{Bin}_R(S)$ is an abelian subgroup of $R$ under $+$ and $\mathrm{Bin}_R(S)$ is an $S$-module under multiplication in $R$.

Let $f: R \times R \to S$ be an $R_S$-Pascal map, and let $S$ be a ring with domain of commutative operators $\Sigma$. A map $B_{n}: \mathrm{Bin}_G(S) \times \Sigma^2 \to S$ is said to be an $f$-Binomial map if it satisfies the equation $$B_n(f)(\alpha, \beta) = \sum_{k = 0}^nf(n\cdot 1_R,k \cdot 1_R)\alpha^{k}\beta^{n-k}$$ for fixed $n \in \mathbb{N}$.

If $T$ denotes the collection of all $f$-Binomial maps, then $(T,+)$ is an abelian group under addition in $S$.

$\textbf{Question:}$ When does $B_n(f)$ admit a closed form or at least admit a type of recursion? If so, what algebraic structure does $T$ admit?

$\textbf{Motivation:}$ I came across these maps after attempting to answer the questions on the properties of the Bhargava factorial, including when there exists a Stirling asymptotic formula for the factorial and the question of the existence of some function of bounded variation over $(0, \infty)$ whose partial moments generate the Bhargava factorial; answers to these question may lead to a generalization of a Riemann zeta-function using the Bhargava factorial via an integral representation. $R$-Pascal maps came up when it came to attempting to "regulate" the Bhargava factorial by determining when its binomial coefficient counterpart satisfies the Pascal identity under ordinary integer addition, and when some "special" addition needs to be realized for Pascal's identity to be satisfied.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ What is a multiplicative identity in an additive group? $\endgroup$
    – darij grinberg
    Commented Mar 1, 2019 at 2:38
  • $\begingroup$ And is $1_R := 1_G$ ? $\endgroup$
    – darij grinberg
    Commented Mar 1, 2019 at 2:38
  • $\begingroup$ How is the norm on $P$ defined in general? $\endgroup$
    – darij grinberg
    Commented Mar 1, 2019 at 2:39
  • $\begingroup$ How is $\Sigma$ defined? $\endgroup$
    – darij grinberg
    Commented Mar 1, 2019 at 2:40
  • $\begingroup$ The way I understand you, the Pascal recurrence (+ induction on $a$) forces your Pascal map $f$ to satisfy $f\left(a, b\right) = \sum\limits_{i=0}^a \dbinom{a}{i} f\left(0, b-i\right)$ for all $a, b \in \mathbb{N}$ (where we understand any integer $m$ inside an argument of $f$ to mean the corresponding element $m \cdot 1_G$ of $G$). The symmetry axiom should restrict it even further. But you cannot mean the two axioms as you stated them. They would not even hold for $\dbinom{n}{k}$ (just set $g_1 = g_2 = 0$ in the recurrence axiom: it would yield $1 = 0 + 0$). $\endgroup$
    – darij grinberg
    Commented Mar 1, 2019 at 2:44

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .